RD Sharma Class 12 Exercise 3.14 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.14 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:49 AM IST

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.14

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Inverse Trigonometric Functions Exercise 3.14 Question 1(i)

Answer: $\frac{-7}{17}$
Hints: First we will solve for $2 \tan ^{-1} \frac{1}{5}$ and we use the formula
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right),-1<x<1$
Given: $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$
Explanation:
$\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$ ..............(1)
Let us first solve for $2 \tan ^{-1} \frac{1}{5}$
$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}$ $\left[\because-1<\frac{1}{5}<1\right]$
$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right\}$
$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{\left(\frac{24}{25}\right)}\right\}$
$=\tan ^{-1}\left\{\frac{2}{5} \times \frac{25}{24}\right\}$
$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{5}{12}\right)$ ................(2)
Now, we know that $\tan \frac{\pi}{4}=1$
$\frac{\pi}{4}=\tan ^{-1}(1)$
Now from equation (1), (2) and (3)
We get $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12} \times 1}\right)\right\}$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \\ \text { forxy }>-1 \end{array}\right]$
$=\tan \left\{\tan ^{-1}\left(\frac{\frac{-7}{12}}{\frac{17}{12}}\right)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{-7}{12} \times \frac{12}{17}\right)\right\}$
$=\tan \left\{\tan ^{-1}\left(\frac{-7}{17}\right)\right\}$ $\left[\because \tan \left(\tan ^{-1} \theta\right)=\theta\right]$ $=\frac{-7}{17}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question

Answer: $\frac{4-\sqrt{7}}{3}$
Hints: First we will convert $\sin ^{-1}\left(\frac{3}{4}\right)$ into $\tan ^{-1}$
Given: $\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right)$
Explanation:
Let $\sin ^{-1}\left(\frac{3}{4}\right)=\theta$ ......(1)
$\sin \theta=\frac{3}{4}=\frac{A B}{A C}$
In $\triangle A B C$
$A B^{2}+B C^{2}=A C^{2}$ (By using Pythagoras theorem)
$\begin{aligned} &3^{2}+B C^{2}=4^{2} \\ &9+B C^{2}=16 \\ &B C^{2}=16-9 \\ &B C^{2}=7 \\ &B C=\sqrt{7} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{B C}{A C} \\ &\cos \theta=\frac{\sqrt{7}}{4} \end{aligned}$
Now
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\frac{\sqrt{7}}{4}}{1+\frac{\sqrt{7}}{4}}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{4-\sqrt{7}}{4}}{\frac{4+\sqrt{7}}{4}}}$
$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$
$\frac{\theta}{2}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$
$\frac{1}{2} \sin ^{-1} \frac{3}{4}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$ from equation (1)
Now, $\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)$
$=\tan \left(\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)\right)$ from equation (2)
$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$
On rationalizing we get,
$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}} \times \frac{4-\sqrt{7}}{4-\sqrt{7}}}$
$=\sqrt{\frac{(4-\sqrt{7})^{2}}{16-7}}$
$\begin{aligned} &=\sqrt{\frac{(4-\sqrt{7})^{2}}{9}} \\ &=\frac{(4-\sqrt{7})}{3} \end{aligned}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 3

Answer: $\frac{1}{\sqrt{10}}$
Hint: First we will convert $\cos ^{-1} \frac{4}{5}$ into $\sin ^{-1}$
Given: $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$
Explanation:
Let $\cos ^{-1} \frac{4}{5}=\theta$
$\cos \theta=\frac{4}{5}$
we know that
$\begin{aligned} &\cos \theta=1-2 \sin ^{2} \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\sin ^{2} \frac{\theta}{2}=\frac{1-\cos \theta}{2} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \end{aligned}$
$\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\frac{4}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{\frac{1}{5}}{2}} \\ &\sin \frac{\theta}{2}=\sqrt{\frac{1}{10}} \\ &\sin \frac{\theta}{2}=\frac{1}{\sqrt{10}} \\ &\frac{\theta}{2}=\sin ^{-1} \frac{1}{\sqrt{10}} \end{aligned}$ from equation (1)
$\frac{1}{2} \cos ^{-1} \frac{4}{5}=\sin ^{-1} \frac{1}{\sqrt{10}}$ ......(2)
Now
$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ from equation (2)
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{10}}\right)$ $\left[\because \sin \left[\sin ^{-1}(\theta)\right]=\theta\right]$
$=\frac{1}{\sqrt{10}}$

Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 4

Answer: $\frac{37}{26}$
Hints: we will convert $2 \tan ^{-1} \frac{2}{3}$ into $\sin ^{-1}$ and $\tan ^{-1} \sqrt{3}$ into $\cos ^{-1}$
Given: $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$
Solution:
$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$ ..................(1)
First we will solve for $2 \tan ^{-1} \frac{2}{3}$
We know that $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad-1<x<1$
$\tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{2 \times \frac{2}{3}}{1+\left(\frac{2}{3}\right)^{2}}\right)$ $-1<\frac{2}{3}<1$
$\begin{aligned} &=\sin ^{-1}\left(\frac{\frac{4}{3}}{1+\frac{4}{9}}\right) \\ &=\sin ^{-1}\left(\frac{\frac{4}{3}}{\frac{13}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{3} \times \frac{9}{13}\right) \end{aligned}$
$2 \tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{12}{13}\right)$ .................(2)
Now Let $\tan ^{-1} \sqrt{3}=\theta$ .................(3)
$\tan \theta=\frac{\sqrt{3}}{1}=\frac{P}{B}$
By Pythagoras theorem:
$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &=\sqrt{3}^{2}+1^{2} \\ &=3+1 \\ &=4 \\ &H=\sqrt{4} \\ &H=2 \\ &\cos \theta=\frac{B}{H} \\ &\cos \theta=\frac{1}{2} \end{aligned}$
$\theta=\cos ^{-1}\left(\frac{1}{2}\right)$ from equation.........(2)
$\tan ^{-1} \sqrt{3}=\cos ^{-1}\left(\frac{1}{2}\right)$ ..................(4)
Now from equation (1)
$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$
Putting the value of $2 \tan ^{-1} \frac{2}{3}$ and $\tan ^{-1} \sqrt{3}$ from equations (2) and (4) respectively
$\begin{aligned} &=\sin \left(\sin ^{-1} \frac{12}{13}\right)+\cos \left(\cos ^{-1} \frac{1}{2}\right) \\ &=\frac{12}{13}+\frac{1}{2} \\ &=\frac{24+13}{26} \\ &=\frac{37}{26} \end{aligned}$

Inverse Trigonometric Functions Exercise 3.14 Question 2

Answer: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Hints: First we will convert $\sin ^{-1} \frac{3}{5}$ into $\tan ^{-1}$ and after that we use the formula of $2 \tan ^{-1} x$
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \quad-1<x<1$
Given: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
Explanation:
$2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$
First we will solve for $\sin ^{-1} \frac{3}{5}$
let $\sin ^{-1} \frac{3}{5}=\theta$ ........................(1)
$\sin \theta=\frac{3}{5}=\frac{P}{H}$
By Pythagoras theorem
$\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$
$\tan \theta=\frac{P}{B}$
$\tan \theta=\frac{3}{4} \quad \theta=\tan ^{-1} \frac{3}{4}$
From Equation (1)
$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ ..............(2)
Now L.H.S:
$2 \sin ^{-1} \frac{3}{5}$
$=2 \tan ^{-1} \frac{3}{4}$ from equation (2)
$=\tan ^{-1}\left[\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right]$ $\left[-1<\frac{3}{4}<1\right.$
$=\tan ^{-1}\left[\frac{\frac{3}{2}}{1-\frac{9}{16}}\right]$
$=\tan ^{-1}\left[\frac{\frac{3}{2}}{\frac{7}{16}}\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{3}{2} \times \frac{16}{7}\right] \\ &=\tan ^{-1}\left[\frac{24}{7}\right] \\ &=R \cdot H . S \end{aligned}$
Hence $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 2

Answer: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$
Hints: First we will solve for $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$ then we will convert it into $\cos ^{-1} \text { and } \sin ^{-1}$ respectively.
Given:
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$
First we will solve for $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{1-\frac{1}{18}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \\ &=\tan ^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) \end{aligned}$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1} \frac{1}{2}$
$=\frac{1}{2} \times 2 \tan ^{-1} \frac{1}{2}$ (On multiplying and dividing by 2)
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \times \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right) \quad\left(\begin{array}{l} \because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ 0 \leq x<\infty \end{array}\right)$
$\begin{aligned} &=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right) \\ &=\frac{1}{2} \cos ^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right) \end{aligned}$
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$ .......................(1)
Now,let $\cos ^{-1} \frac{3}{5}=\theta$ ....................(2)
$\begin{aligned} &\cos \theta=\frac{3}{5} \\ &\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta} \end{aligned}$ $\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$
$\begin{aligned} \sin \theta &=\sqrt{1-\cos ^{2} \theta} \\ \sin \theta &=\sqrt{1-\left(\frac{3}{5}\right)^{2}} \\ &=\sqrt{1-\frac{9}{25}} \\ &=\sqrt{\frac{25-9}{25}} \\ &=\sqrt{\frac{16}{25}} \end{aligned}$
$\begin{aligned} &\sin \theta=\frac{4}{5} \\ &\theta=\sin ^{-1} \frac{4}{5} \end{aligned}$
from equation (2)
$\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}$ .........(3)
From equation (1) and (2) we get
$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 3

Answer: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$
Hint: First we will multiply in numerator and denominator by 2 in L.H.S so that we can use the formula of $2 \tan ^{-1} x$
Given: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$
Explanation:
L.H.S: $\tan ^{-1} \frac{2}{3}$
On multiply in numerator and denominator by 2.
$=\frac{1}{2} \times 2 \tan ^{-1} \frac{2}{3}$
$=\frac{1}{2} \times \tan ^{-1}\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{2}{3}<1 \end{array}\right]$
$\begin{aligned} &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{1-\frac{4}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{\frac{5}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right) \\ &=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \end{aligned}$
Hence it is proved.

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4

Answer: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$
Hints: First we will use the formula of $2 \tan ^{-1} x$ .
Given: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$
Explanation:
$\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{3}<1 \end{array}\right]$
L.H.S: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) \end{aligned}$
$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{3}{28}<1 \end{array}\right]$
$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 5

Answer: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$
Hints: First we will convert $2 \tan ^{-1}\left(\frac{1}{3}\right)$ in $\sin ^{-1}$
Given: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$
Explanation:
L.H.S: $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right]$
$=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^{2}}\right)$ $\left[\begin{array}{c} -1 \leq x \leq 1 \\ -1 \leq \frac{1}{3} \leq 1 \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{\frac{10}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2}{3} \times \frac{9}{10}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{3}{5}\right) \end{aligned}$
$\left[\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right\}\right]$
$\text { [if } \left.x, y \geq 0 \text { and } x^{2}+y^{2} \leq 1\right]$
$\left[\begin{array}{l} \frac{4}{5}, \frac{3}{5} \geq 0,\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \\ \frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1 \leq 1 \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}\left\{\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right\} \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}+\frac{3}{5} \sqrt{1-\frac{16}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{\frac{16}{25}}+\frac{3}{5} \sqrt{\frac{9}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \times \frac{4}{5}+\frac{3}{5} \times \frac{3}{5}\right) \\ &=\sin ^{-1}\left(\frac{16}{25}+\frac{9}{25}\right) \\ &=\sin ^{-1}\left(\frac{25}{25}\right) \end{aligned}$
$\left[\begin{array}{l} \because \sin \frac{\pi}{2}=1 \\ \sin ^{-1}(1)=\frac{\pi}{2} \end{array}\right]$
$\begin{aligned} &=\sin ^{-1}(1) \\ &=\frac{\pi}{2} \end{aligned}$
Hence it is proved that $\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 6

Answer: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$
Hint: First we will convert $\sin ^{-1}\left(\frac{3}{5}\right)$ into $\tan ^{-1}$ .
Given: $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$
Explanation:
Let us solve for $\sin ^{-1}\left(\frac{3}{5}\right)$
Let $\sin ^{-1}\left(\frac{3}{5}\right)=\theta$ ............(1)
$\begin{aligned} &\sin \theta=\left(\frac{3}{5}\right)=\frac{P}{H} \\ &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}$
Now,
$\begin{aligned} &\tan \theta=\frac{P}{B} \\ &\tan \theta=\frac{3}{4} \\ &\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}$
From equation (1)
$\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$ .............(2)
L.H.S:
$2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$
from Equation (2)
$=2 \tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)-\tan ^{-1}\left(\frac{17}{31}\right)$ $\left[\begin{array}{c} -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1}\left(\frac{17}{31}\right) \end{aligned}$
$\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} x \cdot y>-1 \\ \frac{24}{7} \times \frac{17}{31}=\frac{408}{217}>-1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{\frac{625}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$
$\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$
Hence it is Prove that $2 \sin ^{-1}\left(\frac{3}{5}\right)-\tan ^{-1}\left(\frac{17}{31}\right)=\frac{\pi}{4}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7 .
Answer:

$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$
Hint: First we will solve for $2 \tan ^{-1} \frac{1}{5}$
Given: $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$
Explanation:
L.H.S:
$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$
$=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{5}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \end{aligned}$
$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{5}{12} \times \frac{1}{8}=\frac{5}{96}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{10+3}{24}}{1-\frac{5}{96}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right) \\ &=\tan ^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) \\ &=\tan ^{-1}\left(\frac{4}{7}\right) \end{aligned}$
Hence it is proved that $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$

Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 8 .

Answer:

$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$
Hint: First we will solve for $2 \tan ^{-1} \frac{3}{4}$
Given: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$
Explanation:
L.H.S:
$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{3}{4}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1} \frac{17}{31} \end{aligned}$
$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y>-1 \\ \frac{24}{7} \times \frac{17}{31}>-1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that
$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 9.

Answer: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
Hint: First we will solve for
Given: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
Explanation:
$2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right)$ $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{1}{1-\left(\frac{1}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{1}{3}{4}\right) \end{aligned}$
$2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{4}{3}\right)$ ...................(1)
L.H.S:
$2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$
From equation (1)
$=\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$ $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{4}{3} \times \frac{1}{7}=\frac{4}{21}<1 \end{array}\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) \\ &=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right) \\ &=\tan ^{-1}\left(\frac{31}{17}\right) \end{aligned}$
Hence it is proved that $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 10.

Answer: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$
Hints: First we will solve for $4 \tan ^{-1} \frac{1}{5}$
Split into $2 \times 2 \tan ^{-1} \frac{1}{5}$ and solve it
Given: $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$
Explanation: Let us first solve

......(1)
Now
L.H.S:
$\begin{aligned} &4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239} \\ &=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\ &=\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right) \end{aligned}$ $\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28680-119}{28441}}{\frac{28441+120}{28441}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{28561}{28441}}{\frac{28561}{28441}}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
Hence it is proved that $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$

Inverse Trigonomeric Functions Exercise 3.14 Question 3 .

Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}} \text { then } x=\frac{a-b}{1+a b}$
Hints: First we will convert whole L.H.S part in $\tan ^{-1}$
Given: $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$
Explanation:
As we know that
$\begin{aligned} &2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ &2 \tan ^{1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}$
Now $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$
$\begin{aligned} &2 \tan ^{-1} a-2 \tan ^{-1} b=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2\left(\tan ^{-1} a-\tan ^{-1} b\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x \\ &\frac{a-b}{1+a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a-b}{1+a b}=x \\ &x=\frac{a-b}{1+a b} \end{aligned}$
Hence it is proved that $x=\frac{a-b}{1+a b}$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1 .

Answer: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$
Hint: First we will convert $\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$ into $\tan ^{-1}$
Given: $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$
Explanation:
L.H.S:
$\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)$
$=\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$ $\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$
$=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right]$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right] \\ &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right] \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$
$=\tan ^{-1}(\infty)$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$
$=\frac{\pi}{2}$
Hence it is Proved that $\tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 2.

Answer: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$
Hint: First we will convert $\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}$ into $\tan ^{-1}$
Given: $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$
Explanation:
L.H.S:
$\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}$
$=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+2 \tan ^{-1} x\right\}$ $\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x\right]$
$=\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\tan ^{-1} \frac{2 x}{1-x^{2}}\right\}$ $\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$
$\begin{aligned} &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-\left(\frac{1-x^{2}}{2 x}\right)\left(\frac{2 x}{1-x^{2}}\right)}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right\} \\ &=\sin \left\{\tan ^{-1} \frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right\} \end{aligned}$ $\left(\frac{a}{0}=\infty\right)$
$=\sin \left\{\tan ^{-1}(\infty)\right\}$ $\left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)$
$\begin{aligned} &=\sin \left(\frac{\pi}{2}\right) \\ &=1 \end{aligned}$
Hence it is proved that $\sin \left\{\tan ^{-1} \frac{1-x^{2}}{2 x}+\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}=1$

Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 3.

Answer: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$
Hints: We will first convert $2 x \sqrt{1-x^{2}}$ into sin
Given: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$
$x=\sin \theta$
Let $\sqrt{1-x^{2}}=\sqrt{1-\sin ^{2} \theta}=\sqrt{\cos ^{2} \theta}=\cos \theta$ $\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right]$
Now
$\begin{aligned} &\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \\ &=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^{2} \theta}\right) \\ &=\sin ^{-1}(2 \sin \theta \cos \theta) \end{aligned}$ $[2 \sin \theta \cos \theta=\sin 2 \theta]$
$\begin{aligned} &=2 \theta \\ &=2 \sin ^{-1} x \end{aligned}$ $\left[\begin{array}{l} \because \sin ^{-1}(\sin \theta)=\theta \\ x=\sin \theta \\ \theta=\sin ^{-1} x \end{array}\right]$
Hence it is proved that $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$

Inverse Trigonomeric Functions Exercise 3.14 Question 5.

Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$ to prove $x=\frac{a+b}{1-a b}$
Hints: First we will convert L.H.S of the question in $\tan ^{-1}$
Given: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$
Explanation:
$\begin{aligned} &\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \\ &\because 2 \tan ^{-1} a=\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right) \\ &\therefore \sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \end{aligned}$
$\begin{aligned} &2 \tan ^{-1} a+2 \tan ^{-1} b=2 \tan ^{-1} x \\ &2\left(\tan ^{-1} a+\tan ^{-1} b\right)=2 \tan ^{-1} x \\ &\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} x \end{aligned}$
$\tan ^{-1}\left(\frac{a+b}{1-a b}\right)=\tan ^{-1} x$ $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &\frac{a+b}{1-a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a+b}{1-a b}=x \end{aligned}$
Hence it is proved that $x=\frac{a+b}{1-a b}$

Inverse Trigonomeric Functions Exercise 3.14 Question 6.

Answer: $\pi$
Hint: First we will convert $\sin ^{-1} \frac{2 x}{1+x^{2}}$ into $\tan ^{-1}$
Given: $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ for $x \geq 1$
Explanation:
$2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ $\left[\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right]$
$\begin{aligned} &=2 \tan ^{-1} x+2 \tan ^{-1} x \\ &=4 \tan ^{-1} x \quad(x \geq 1) \end{aligned}$
Let $x=1$
$\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$
$\begin{aligned} &=4 \tan ^{-1}(1) \\ &=4 \times \frac{\pi}{4} \\ &=\pi \end{aligned}$
Hence the constant value for $x \geq 1$ is $\pi$

Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 1.

Answer:

$\frac{\pi}{4}$
Given: Given that $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
Hint: First we solve $2 \sin ^{-1} \frac{1}{2}$ Since $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
Solution: We have,
$\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$
$=\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}$ $\left[\sin ^{-1} \frac{1}{2}=\frac{\pi}{6}\right]$
$\begin{aligned} &=\tan ^{-1}\left\{2 \cos \frac{\pi}{3}\right\} \\ &=\tan ^{-1}\left\{2 \times \frac{1}{2}\right\} \end{aligned}$ $\left[\cos \frac{\pi}{3}=\frac{1}{2}\right]$
$\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$
This is the required solution.

Inverse Trigonomeric Functions Exercise 3.14 Question 7 Sub Question 2.

Answer: 0
Given: $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right),|x| \geq 1$
Hint: Since $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$ applying it.
Solution: We have,
$\begin{aligned} &\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right) \\ &=\cos \left(\frac{\pi}{2}\right) \end{aligned}$ $\left[\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right]$
=0 $\left[\cos \frac{\pi}{2}=0\right]$
Hence, $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=0$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 1.

Answer: $x=\frac{-461}{9}$
Given: $\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
Hint: Use formula
$\text { 1. } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\text { 2. } 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Solution: We have,
$\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
$\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{4}+\frac{5}{12}}{1-\frac{1}{4} \times \frac{5}{12}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{8}{12}}{\frac{43}{48}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{32}{43}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \end{aligned}$
$\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1-\frac{32}{43} \times \frac{1}{6}}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$ $\left[\tan ^{-1} \frac{\pi}{4}=1\right]$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{258}}{\frac{226}{258}}\right)+\tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\tan ^{-1}(1)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{235}{226}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}(1) \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\left(\frac{235}{226}\right) \times \frac{1}{x}}\right)=\tan ^{-1}(1), \frac{235}{226} x<1 \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{235 x+226}{226 x-235}=1, x>\frac{235}{226} \\ &\Rightarrow 235 x+226=226 x-235, x>\frac{235}{226} \\ &\Rightarrow 235 x-226 x=-235-226 \\ &\Rightarrow 9 x=-461 \\ &\Rightarrow x=\frac{-461}{9} \end{aligned}$
Hence $x=\frac{-461}{9}$ is required answer

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2.

Answer:

$x=\frac{1}{\sqrt{3}}$
Given: $3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
Hint: Using formula since,
$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution: We have,
$3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
We know that $2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\begin{aligned} &\Rightarrow 3\left(2 \tan ^{-1} x\right)-4\left(2 \tan ^{-1} x\right)+2\left(2 \tan ^{-1} x\right)=\frac{\pi}{3} \\ &\Rightarrow 6 \tan ^{-1} x-8 \tan ^{-1} x+4 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \end{aligned}$
$\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ $\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\right]$
$\Rightarrow x=\frac{1}{\sqrt{3}}$
This is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 3.

Answer: $x=\frac{1}{\sqrt{3}}$
Given: $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}, x>0$
Hint: Use
$\text { 1. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \text { and }$
$\text { 2. } \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution: We have,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}$
We know that,
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$ and convert $\cot ^{-1}$ into $\tan ^{-1}$
$\Rightarrow 2 \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$ $\left[\cot \theta=\frac{1}{\tan \theta}\right]$
$\Rightarrow 2 \tan ^{-1} x+2 \tan ^{-1} x=\frac{2 \pi}{3}$ $\left[\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow 4 \tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{2 \pi}{12} \\ &\Rightarrow x=\tan \frac{\pi}{6} \end{aligned}$ $\left[\tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}\right]$
$\Rightarrow x=\frac{1}{\sqrt{3}}$ $\left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 4.

Answer: $\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$
Given: $2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x), x \neq \frac{\pi}{2}$
Hint: Using $2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
Solution: We have,
$2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x)$
We know that
$2 \tan ^{-1}(x)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}(2 \sec x) \\ &\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=2 \sec x \end{aligned}$ $\left[1-\sin ^{2} x=\cos ^{2} x\right]$
$\Rightarrow \frac{\sin x}{\cos x \cos x}=\frac{1}{\cos x}$ $\left[\sec x=\frac{1}{\cos x}\right]$
$\Rightarrow \tan x=1$ $\left[\frac{\sin x}{\cos x}=\tan x\right]$
$\begin{aligned} &\Rightarrow x=\tan ^{-1}(1) \\ &\Rightarrow x=\frac{\pi}{4} \end{aligned}$ $\left[\tan ^{-1}(1)=\frac{\pi}{4}\right]$
Hence the principal value of $x=n \pi+\frac{\pi}{4}, n \in Z$

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 5.

Answer: $x=\sqrt{3}$
Given: $\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$
$\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution:
We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$
$\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
$\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \end{aligned}$
$\Rightarrow x=\sqrt{3}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$
Hence $x=\sqrt{3}$ is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 6.

Answer: $x=\sqrt{3}$
Given: $\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
Hint: Using formula
$\text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x$
$\text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x$
Solution:
We have,
$\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$
We know that,
$\begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}$ $[\cos (-\theta)=\pi-\cos \theta]$
$\begin{aligned} &\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1} x=\frac{2 \pi}{3} \end{aligned}$
$\left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]$
$\begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \\ &\Rightarrow x=\sqrt{3} \end{aligned}$ $\left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]$
Hence $x=\sqrt{3}$ is required solution.

Inverse Trigonometric Function Exercise 3.14 Question 9.

Answer: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
Given: $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
Hint: Using $2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution: Let us assume
$\begin{aligned} \text { L.H.S } &=2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right] \\ &=\cos ^{-1}\left[\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right)^{2}}\right] \end{aligned}$ $\left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$
$=\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2} \frac{\theta}{2}}\right]$
$\begin{aligned} &=\cos ^{-1}\left[\frac{a+b-(a-b) \tan ^{2} \frac{\theta}{2}}{a+b+(a-b) \tan ^{2} \frac{\theta}{2}}\right] \\ &=\cos ^{-1}\left[\frac{a\left(1-\tan ^{2} \frac{\theta}{2}\right)+b\left(1+\tan ^{2} \frac{\theta}{2}\right)}{a\left(1+\tan ^{2} \frac{\theta}{2}\right)+b\left(1-\tan ^{2} \frac{\theta}{2}\right)}\right] \end{aligned}$
Dividing numerator and denominator by $\left(1+\tan ^{2} \frac{\theta}{2}\right)$ we get
$=\cos ^{-1}\left(\frac{a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+b}{a+b\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)}\right]$
$=\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right]$ $\left[\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\cos \theta\right]$
$= R.H.S$
Hence $2 \tan ^{-1}\left[\sqrt{\frac{a-2}{a+2}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$

Inverse Trigonometric Functions Exercise 3.14 Question 10.

Answer: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
Given: $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$ where $\alpha=a x-b y \: \: \& \: \: \beta=a y+b x$
Hint: Use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Solution: Let us assume
L.H.S: $=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\left(\frac{2 a b}{a^{2}-b^{2}}\right)\left(\frac{2 x y}{x^{2}-y^{2}}\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 a b x^{2}-2 a b y^{2}+2 x y a^{2}-2 x y b^{2}}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2\left(a b x^{2}+x y a^{2}-a b y^{2}-x y b^{2}\right)}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2[a x(b x+a y)-b y(a y+b x)]}{(a x-b y)^{2}-\left(a^{2} y^{2}+b^{2} x^{2}+2 a b x y\right)}\right] \end{aligned}$
$=\tan ^{-1}\left[\frac{2(b x+a y)(a x-b y)}{(a x-b y)^{2}-(b x+a y)^{2}}\right]$ $\left[(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
$=\tan ^{-1}\left[\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right]$ $[\alpha=a x-b y \& \beta=a y+b x]$
$=R.H.S$
Hence , $\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$

Inverse Trigonomeric Functions Exercise 3.14 Question 11.

Answer: $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
Given: For any a, b, x, y > 0
We have to prove that $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$
where, $\alpha=-a x+b y \& \beta=b x+a y$
Hint: First we will divide numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$ respectively, then use $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Solution: Let us assume
$\begin{aligned} L . H . S &=\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right) \\ &=\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 a b^{2}-a^{3}}{b^{3}}}{\frac{b^{3}-3 a^{2} b}{b^{3}}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 x y^{2}-x^{3}}{y^{3}}}{\frac{y^{3}-3 x^{2} y}{y^{3}}}\right] \end{aligned}$
Dividing numerator and denominator of first function and second function by $b^{3}$ and $y^{3}$ respectively.
$=\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{a}{b}\right)-\left(\frac{a}{b}\right)^{3}}{1-3\left(\frac{a}{b}\right)^{2}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{x}{y}\right)-\left(\frac{x}{y}\right)^{3}}{1-3\left(\frac{x}{y}\right)^{2}}\right]$
We know that, $3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
$\Rightarrow \mathrm{LH} . \mathrm{S}=\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right]+\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]$
$\begin{aligned} &=2 \tan ^{-1}\left(\frac{a}{b}\right)+2 \tan ^{-1}\left(\frac{x}{y}\right) \\ &=2\left[\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{x}{y}\right)\right] \end{aligned}$
$=2 \tan ^{-1}\left[\frac{\frac{a}{b}+\frac{x}{y}}{1-\frac{a}{b} \times \frac{x}{y}}\right]$ $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$
$\begin{aligned} &=2 \tan ^{-1}\left[\frac{a y+b x}{b y-a x}\right] \\ &=2 \tan ^{-1}\left(\frac{\beta}{\alpha}\right) \end{aligned}$ $[b y-a x=\alpha \& \beta=a y+b x]$
$=\tan ^{-1}\left(\frac{2 \times \frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{2}}\right)$ $\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$
$\begin{aligned} &=\tan ^{-1}\left(\frac{2 \beta}{\alpha} \times \frac{\alpha^{2}}{\alpha^{2}-\beta^{2}}\right) \\ &=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) \\ &=\text { R.H.S } \end{aligned}$
Hence, $\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)$ Where $\alpha=-a x+b y \& \beta=b x+a y$

RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 involves inverse trigonometric problems related to cosecant, secant, cosine, tangent functions. Solving these problems will make a student grasp better and perform well in exams.

In calculus, Inverse Trigonometric Functions are used to find various integrals. In addition to mathematics, inverse trigonometric functions are applied in science and engineering. Students will learn about the domain and range restrictions of trigonometric functions that ensure the existence of their inverses and how to observe their behavior using graphical representations and examples.

The specific exercise 3.14 has 31 questions, including subparts, and now a student can practice a lot and gain knowledge about the topic. In addition, this exercise will help in overall understanding of the chapter. RD Sharma Class 12 Solutions Chapter 3 Exercise 3.14 has practice questions which are to be solved and are important for exam point of view. The concepts are now easily understood by the students.

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